To solve the quadratic equation 6x² + 13x + 5 = 0, we can use the quadratic formula, which is given by:
x = \frac{-b \pm \sqrt{b² – 4ac}}{2a}
In this equation, the variables are:
- a = 6
- b = 13
- c = 5
First, we need to calculate the discriminant:
b² – 4ac = 13² – 4(6)(5)
= 169 – 120
= 49
Since the discriminant is positive, we can expect two distinct real solutions.
Now, substituting the values of a, b, and c into the quadratic formula:
x = \frac{-13 \pm \sqrt{49}}{2(6)}
x = \frac{-13 \pm 7}{12}
This gives us two potential solutions:
- x₁ = \frac{-13 + 7}{12} = \frac{-6}{12} = -\frac{1}{2}
- x₂ = \frac{-13 – 7}{12} = \frac{-20}{12} = -\frac{5}{3}
Thus, the solution set of the equation 6x² + 13x + 5 = 0 is:
{ -\frac{1}{2}, -\frac{5}{3} }